Worksheet: Noetherian Rings and Modules

Worksheet: Noetherian rings and modules

Deﬁnition: An R-module M is called Noetherian if every ascending chain of R- submodules M1 ⊆ M2 ⊆ M3 ⊆ · · · of M is eventually stable, i.e., there exists some index N for which Mn = Mn+1 for all n ≥ N. We say R is a Noetherian ring, or simply say that R is Noetherian, if it is Noetherian when regarded as a module over itself.

Warmup. (1) Explain what it means for R to be Noetherian in terms of ascending chains of ideals. (2) Verify that ﬁelds and PIDs are Noetherian. (3) Must a UFD be Noetherian? Hint: Observe that the ascending union of UFDs is a UFD. (4) Explain why, if R is Noetherian, then so is R/I for every ideal I ⊆ R. (5) Explain why, R is Noetherian, then so is U −1R for every multiplicative set U ⊆ R. Hint: Recall (or reprove, if you forgot) that every ideal of U −1R is the expansion of its contraction to R.

Examples and nonexamples. By the end of this worksheet, we will see how to construct many interesting examples of Noetherian rings. For now, we consider some simple examples. (1) Provide an example of a nonzero ring R and a nonzero R-module M such that (a) R and M are both Noetherian; (b) R is Noetherian, but M is not; (c) R is not Noetherian, but M is; (d) neither R nor M is Noetherian. (2) Let R be the R-algebra consisting of all continuous functions f : [0, 1] → R, with addition and multiplication of functions deﬁned pointwise. Prove that R is not a Noetherian ring. Hint: Fix a proper closed subinterval I of the unit interval, and consider the set of all f ∈ R vanishing at each point of I.

Some characterizations. Verify that the following conditions on a ring R are equivalent. (a) R is Noetherian. (b) Every non-empty family of ideals {Iα}α∈A of R has a maximal element, i.e., there exists β ∈ A such that if α ∈ A and Iβ ⊆ Iα, then Iβ = Iα.( !!)We do not require that any member of the family be contained in such a maximal element, besides the element itself (!!) (c) Every ascending chain of finitely generated ideals of R is eventually stable. (d) For every ideal I ⊆ R, and for every subset G ⊆ R with I = hGi, there exists a finite subset G◦ ⊆ G such that I = hG◦i. (e) Every ideal of R is finitely generated.

Noetherian modules. Convince yourself that the following conditions on an R-module M are equivalent. You don’t need to write a solution, unless something is not obvious to you. (a) M is a Noetherian R-module. (b) Every non-empty family of submodules {Mα}α∈A of M has a maximal element, i.e., there exists β ∈ A such that if α ∈ A and Mβ ⊆ Mα, then Mβ = Mα. (c) Every ascending chain of finitely generated submodules of M is eventually stable. (d) For every submodule N ⊆ M, and for every subset G ⊆ M that generates N, there exists a finite subset G◦ ⊆ G such that also generates N. (e) Every submodule of M, including M itself, is finitely generated. 1 2

Connections with short exact sequences. Our goal here is to prove the following.

φ Proposition: If 0 → N −→θ M −→ Q → 0 is a short exact sequence of R-modules, then M is a Noetherian R-module if and only if both N and Q are Noetherian R-modules.

(1) Prove the following Lemma: In the above notation, if M0 ⊆ M1 ⊆ M are submodules such that M0 ∩ θ(N) = M1 ∩ θ(N) and φ(M0) = φ(M1), then M0 = M1. (2) Prove the Proposition. Hint: Apply the Lemma for the “ ⇐= ” implication. (3) Prove the following corollary, which provides a simpliﬁed characterization of when a given R-module is Noetherian in the special case that the ring R itself is Noetherian.

Corollary: If R is Noetherian, then an R-module M is Noetherian if and only if M is finitely generated. In particular, if R is Noetherian, then every submodule of a finitely generated module is also finitely generated.

Hint: For the “ ⇐= ” implication, first prove that the finitely generated free module n Fn = ⊕i=1R is Noetherian for each n ≥ 1. If M is a finitely generated R-module, can you fit it into a SES that involves Fn for some n? (4) Show that the last statement in the Corollary can fail by when R is not Noetherian. Another characterization. Our goal in this problem is to prove a theorem of Cohen, which tells us that one can relax the last condition in our list of characterizations of Noetherian rings.

Theorem: A ring R is Noetherian if and only if each prime ideal of R is ﬁnitely generated.

(1) First, verify that if I ⊆ R is an ideal, and x ∈ R, then (I : x) := {r ∈ R : rx ∈ I} is an ideal of R. (2) We prove the “ ⇐= ” implication by contradiction. Suppose that this implication is false. (a) Let X be the collection all ideals of I of R that are not finitely generated. Explain why X is non-empty, and why X contains a maximal element, i.e., why there exists an ideal in X not properly contained in any other ideal in X. Hint: Review Zorn’s Lemma. (b) Let P ∈ X be a maximal element of X. We will now show that P must be a prime ideal of R. By contradiction, suppose that P is not prime, and fix x, y ∈ R \ P such that xy ∈ P . Explain why we can write hxi + P = hx, p1, . . . , pì with all pi ∈ P . Hint: Why is it that hxi+P is even finitely generated? Once you know this, you know it is generated by finitely many elements that look like ax + p with a ∈ R and p ∈ P . Can you use generators of this form to produce the desired generating set? (c) Set J = (P : x), and explain why P = xJ + hp1, . . . , pì. Hint: To establish the containment “⊆”, use the fact that P ⊆ hxi+P = hx, p1, . . . , pì. (d) Explain why J must be finitely generated. Hint: Have you used y yet? (e) Complete the proof by deriving a contradiction from the last two points. (3) It is now natural to ask where else we can replace ideal with prime ideal in a characterization of a Noetherian ring to obtain a relaxed characterization. For example, consider the Question: If every ascending chain of prime ideals of R is eventually stable, then must R be Noetherian? Show that the answer to this question is “No.” Hint: There are elementary examples. E.g., can you find a ring that is not Noetherian, but has only one prime ideal? 3

Hilbert Basis Theorem. Our goal is to prove the following fundamental result.

Theorem (Hilbert Basis Theorem): If R is a Noetherian ring, then so is R[x].

n (1) Given a nonzero polynomial f(x) = a0 + a1x + ··· anx with each ai ∈ R, and an 6= 0, let LT(f) = an denote the leading term of f. We adopt the convention LT(0) = 0. Given an ideal I of R[x], prove that LT(I) := {LT(f): f ∈ I} is an ideal of R, possibly improper, whether R is Noetherian, or not. (2) For the remainder of this sketch, suppose that R is Noetherian, and that I is an ideal of R[x]. Explain why there exist f1, . . . , f` ∈ I such that LT(I) = hLT(f1),..., LT(f`)i. (3) Let m = max{deg(f1),..., deg(f`)}. Show that N = {g ∈ I : deg(g) ≤ m} ∪ {0} is an R-submodule of R[x], but not an ideal of R[x]. (4) Prove that N is a finitely generated R-module. Hint: As R is assumed to be Noetherian, we know from an earlier Corollary that if we can find a finitely generated (and hence, Noetherian) R-module M containing N, then this will force N to be finitely generated. Seek a natural choice of M that satisfies this condition. (5) Prove that every element f ∈ I can be written in the form h + g with h ∈ hf1, . . . , fì and g ∈ N. Hint: Induce on deg(f); the case when deg(f) ≤ m is trivial. (6) Conclude the proof of the Hilbert Basis Theorem by explaining why I is finitely generated. (7) If R is a ring, then a ring S is said to be of finite type over R if S is a finitely generated R-algebra, and is said to be essentially of finite type over R if it is the localization of a finitely generated R-algebra at some multiplicatively closed set. Observe that almost every explicitly defined ring we’ve seen in this course (the exceptions being rings like the R-algebra of continuous function f : [0, 1] → R that we saw earlier, and polynomial rings with countably many variables) are essentially of finite type over a field, or over a PID. Prove the following Corollary, which shows that each such ring must be Noetherian. Hint: Deduce a generalization of the Hilbert Basis Theorem that involves finitely many variables, and apply some results from the Warmup.

Corollary: Every ring essentially of finite type over a Noetherian ring is also Noe- therian. In particular, every ring essentially of finite type over a field is Noetherian

Some basic applications. Here, we give a hint of the types of results that rely on the Noetherian condition. For each problem, make certain to explicitly state which characterization you use. (1) Let φ : M → M be an R-linear endomorphism of a Noetherian R-module M. Prove that if φ n is onto, then φ must be an automorphism. Hint: Verify that Kn = ker(φ ) is a submodule M for each n. Also, note that each iterate φn : M → M must also be onto. (2) Prove that if R is Noetherian, then every ideal contains a power of its radical.√ Conclude that, given an ideal I of a Noetherian ring R, there exists N such that x ∈ I ⇐⇒ xN ∈ I. (3) Prove that the previous conclusion can fail if R is not Noetherian.